Search results
Results from the WOW.Com Content Network
Then | | + + + + + | | so | | + + + + + | | This shows that the sum of the four integrals (in the middle) is finite if and only if the integral of the absolute value is finite, and the function is Lebesgue integrable only if all the four integrals are finite. So having a finite integral of the absolute value is equivalent to the conditions for ...
If the function f does not have any continuous antiderivative which takes the value zero at the zeros of f (this is the case for the sine and the cosine functions), then sgn(f(x)) ∫ f(x) dx is an antiderivative of f on every interval on which f is not zero, but may be discontinuous at the points where f(x) = 0.
A surface integral generalizes double integrals to integration over a surface (which may be a curved set in space); it can be thought of as the double integral analog of the line integral. The function to be integrated may be a scalar field or a vector field. The value of the surface integral is the sum of the field at all points on the surface.
If |x| 1 and |x| 2 are two absolute values on the same integral domain D, then the two absolute values are equivalent if |x| 1 < 1 if and only if |x| 2 < 1 for all x. If two nontrivial absolute values are equivalent, then for some exponent e we have |x| 1 e = |x| 2 for all x. Raising an absolute value to a power less than 1 results in another ...
The same definition can be used for series = whose terms are not numbers but rather elements of an arbitrary abelian topological group.In that case, instead of using the absolute value, the definition requires the group to have a norm, which is a positive real-valued function ‖ ‖: + on an abelian group (written additively, with identity element 0) such that:
The term "quadrature" is a traditional term for area; the integral is geometrically interpreted as the area under the curve y = x n. Traditionally important cases are y = x 2, the quadrature of the parabola, known in antiquity, and y = 1/x, the quadrature of the hyperbola, whose value is a logarithm.
Sometimes integrals may have two singularities where they are improper. Consider, for example, the function 1/((x + 1) √ x) integrated from 0 to ∞ (shown right). At the lower bound of the integration domain, as x goes to 0 the function goes to ∞, and the upper bound is itself ∞, though the function goes to 0. Thus this is a doubly ...
A different technique, which goes back to Laplace (1812), [3] is the following. Let = =. Since the limits on s as y → ±∞ depend on the sign of x, it simplifies the calculation to use the fact that e −x 2 is an even function, and, therefore, the integral over all real numbers is just twice the integral from zero to infinity.